-
We plot the Feynman digrams of the decay process
$ B_0 \to J/\psi K^0 f_0(a_0)(980) $ involving a triangle loop in Figs. 1 and 2. We observe that particle$ B_0 $ first decays into particles$ J/\psi, \phi, K^0 $ , and then, particle ϕ decays into$ K^0 $ and$ \bar{K}^0 $ ; then,$ K^0 $ and$ \bar{K}^0 $ fuse into$ f_0(a_0) $ , and eventually$ f_0(a_0) $ decays into$ \pi^+\pi^-(\pi^0\eta) $ .Figure 1. Feynman diagrams of the decay process
$B_0 \to $ $ J/\psi K^0 f_0(980)\to J/\psi K^0\pi^+ \pi^-$ involving a triangle loop.Figure 2. Feynman diagrams of the decay process
$B_0 \to $ $ J/\psi K^0 a_0(980)\to J/\psi K^0\pi^0 \eta$ involving a triangle loop.We take Fig. 1 for example to perform the following discussion, since Figs. 1 and 2 have nearly the same amplitude. Now, we want to find the position of TS in the complex-q plane, instead of evaluating the whole amplitude of a Feynman diagram including the triangle loop. Analogously to Refs. [18, 36, 37], we can use
$ \begin{align} q_{\rm on+}=q_{a-} , \end{align} $
(1) where
$q_{\rm on+}$ represents the on shell momentum of the ϕ in the center of mass frame of$ \phi K^0 $ . Meanwhile,$ q_{a-} $ can be obtained by analyzing the singularity structure of the triangle loop;$ q_{a-} $ represents the on-shell$ K^0 $ momentum in the loop, antiparallel to the ϕ.$q_{\rm on+}$ and$ q_{a-} $ are given by$ \begin{align} q_{\rm on+}=\frac{\lambda^{\frac{1}{2}}(s,M_{\phi}^2,M^2_{K^0})}{2\sqrt{s}},\qquad q_{a-}=\gamma(vE_{K^0}-p^*_{K^0})-{\rm i}\epsilon, \end{align} $
(2) where s denotes the squared invariant mass of ϕ and
$ K^0 $ , and$ \lambda(x,y,z)=x^2+y^2+z^2-2xy-2yz-2xz $ is the Kählen function, with definitions$ \begin{aligned}[b] &\ \quad v=\frac{k}{E_{f_0}},\ \qquad\qquad\qquad\qquad\gamma=\frac{1}{\sqrt{1-v^2}}=\frac{E_{f_0}}{m_{f_0}},\\ &E_{K^0}=\frac{m_{K^+_1}^2+m_{K^0}^2-m_{\bar{K}^0}^2}{2m_{f_0}},\;\;\; p^*_{K^0}=\frac{\lambda^{\frac{1}{2}}(m^2_{f_0},m_{K^0}^2,m_{\bar{K}^0}^2)}{2m_{f_0}}. \end{aligned} $
(3) It is easy to realize that
$ E_{K^0} $ and$ p^*_{K^0} $ are the energy and momentum of$ K^0 $ in the center of mass frame of the$ \phi K^0 $ system, and v and γ are the velocity of the$ f_0 $ and Lorentz boost factor. In addition, we can easily obtain$ \begin{align} E_{f_0}=\frac{s+m^2_{f_0}-m^2_{\bar{K}^0}}{2\sqrt{s}},\;\;\; k=\frac{\lambda^{\frac{1}{2}}(s,m_{f_0}^2,m^2_{\bar{K}^0})}{2\sqrt{s}}. \end{align} $
(4) The three intermediate particles must be on the shell. We let the mass of
$ f_0 $ be slightly larger than the sum of the masses of$ K^0 $ and$ \bar{K}^0 $ ; now, we can determine that the mass of$ K^0 $ is$ 497.61\ \rm M\rm eV $ , the mass of$ f_0 $ is slightly larger than the sum of the masses of$ K^0 $ and$ \bar{K}^0 $ , and the mass of$ f_0 $ is$ 991\ \rm{MeV} $ ; additionally, we can find a TS at approximately$ \sqrt{s}=1520\ \rm{MeV} $ . If we use in Eq. (1) complex masses$ (M-{\rm i} \Gamma/2) $ of mesons that include widths of$ K^0 $ and$ \bar{K}^0 $ , the solution of Eq. (1) is then$(1520- 10{\rm i})\ \rm{MeV}$ . This solution implies that the TS has a "width" of$ 20\ \rm{MeV} $ . -
Here, we only focus on the
$ B_0\to J/\psi\phi K^0 $ process. The decay branching ratio has been experimentally measured to be [38]$ \begin{align} {\rm{Br}}(B_0\to J/\psi\phi K^0)=(4.9\pm 1.0)\times10^{-5}. \end{align} $
(5) The differential decay width over the invariant mass distribution
$ J/\psi \phi $ can be written as$ \begin{align} \frac{\rm \Gamma_{B_0\to J/\psi\phi K^0}}{\rm {M_{\rm{inv}}}(J/\psi\phi)}=\frac{1}{(2\pi)^3}\frac{{ \tilde{p}_{\phi}} {p_{K^0}}}{4M_{B_0}^2}\cdot \overline{\sum}|t_{B_0\to J/\psi \phi K^0}|^2, \end{align} $
(6) here,
$ \tilde{p}_{\phi} $ represents the momentum of the$ K^0 $ in the$ \phi K^0 $ rest frame, and$ {p_{K^0}} $ represents the momentum of the$ K^0 $ in the$ B_0 $ rest frame:$ \begin{align} \tilde{p}_{\phi}&=\frac{\lambda^{\frac{1}{2}}(M_{\rm inv}(J/\psi\phi)^2,m_{J/\psi}^2,m_{\phi}^2)}{2M_{\rm inv}(J/\psi\phi)}, \end{align} $
(7) $ \begin{align} p_{K^0}&=\frac{\lambda^{\frac{1}{2}}(M_{B_0}^2,m_{K^0}^2,M_{\rm inv}(J/\psi\phi)^2)}{2M_{B_0}}.\\ \end{align} $
(8) We use the polarization summation formula
$ \begin{align} \sum\limits_{\rm pol}=\epsilon_{\mu}(p)\epsilon_{\nu}(p)=-g_{\mu\nu}+\frac{p_{\mu}p_{\nu}}{m^2}, \end{align} $
(9) and we have
$ \begin{aligned}[b] \overline{\sum_{\rm pol}}|t_{B_0\to J/\psi\phi K^0}|^2&=\mathcal{C}^2 \left( -g_{\mu\nu}+\frac{p_{J/\psi}^\mu p_{J/\psi}^\nu}{m_{J/\psi}^2} \right)\left( -g_{\mu\nu}+\frac{p_{\phi}^\mu p_{\phi}^\nu}{m_{\phi}^2} \right)\\ &=\mathcal{C}^2\left( 2+\frac{(p_{J/\psi} \cdot p_{\phi})^2}{m_{J/\psi}^2 m_{\phi}^2} \right), \end{aligned} $
(10) where
$ \begin{align} p_{\phi}\cdot p_{J/\psi}=\frac{1}{2}(M_{\rm inv}(J/\psi\phi)^2-m_{J/\psi}^2-m_{\phi}^2), \end{align} $
(11) $ \begin{align} \frac{1}{\Gamma_{B_0}}=\frac{{\rm{Br}}(B_0\to J/\psi\phi K^0)}{\int\dfrac{{\rm d} \Gamma_{B_0}}{{\rm d}M_{\rm inv}(J/\psi\phi)}{\rm d}M_{\rm inv}(J/\psi\phi)}. \end{align} $
(12) Then, we can determine the value of the constant
$ \mathcal{C} $ $ \begin{align} \frac{\mathcal{C}^2}{\Gamma_{B_0}}=\frac{{\rm{Br}}(B_0\to J/\psi\phi K^0)}{\int {\rm d}M_{\rm inv}(J/\psi\phi)\dfrac{1}{(2\pi)^3}\dfrac{\tilde{p}_{\phi} p_{K^0}}{4M_{B_0}^2}\cdot\left( 2+\dfrac{(p_{J/\psi} \cdot p_{\phi})^2}{m_{J/\psi}^2 m_{\phi}^2} \right)}, \end{align} $
(13) where the integration is performed from
$ M_{\rm inv}(J/\psi\phi)_{\rm{min}} $ =$ m_{J/\psi}+m_{\phi} $ to$ M_{\rm inv}(J/\psi\phi)_{\rm{max}} $ =$ M_{B_0}-m_{K^0} $ . -
In the previous subsection, we calculated the transition strength of the decay process
$ B_0\to J/\psi \phi K^0 $ . Now, we calculate the contribution of the vertex$ \phi\to K^0\bar{K}^0 $ . We can obtain this$ VPP $ vertex from the chiral invariant Lagrangian with local hidden symmetry given in Refs. [39–42]:$ \begin{align} \mathcal{L}_{VPP}=-{\rm i}g\langle V^{\mu}[P,\partial_{\mu}P]\rangle, \end{align} $
(14) where
$ \langle\rangle $ denotes the trace of the flavor SU(3) matrices, and g represents the coupling in the local hidden gauge.$ \begin{align} g=\frac{M_V}{2f_{\pi}},\qquad M_V=800\ {\rm MeV},\qquad {f_{\pi}}=93\ \rm MeV. \end{align} $
(15) The V and P in Eq. (14) are the vector meson matrix and pseudoscalar meson matrix in the SU(3) group, respectively, which are given by
$ \begin{align} &P=\left( \begin{array}{ccc} \dfrac{\pi^{0}}{\sqrt{2}}+\dfrac{\eta}{\sqrt{3}}+\dfrac{\eta^{\prime}}{\sqrt{6}} & \pi^+ & K^+ \\ \pi^- & -\dfrac{\pi^{0}}{\sqrt{2}}+\dfrac{\eta}{\sqrt{3}}+\dfrac{\eta^{\prime}}{\sqrt{6}} & K^{0} \\ K^- & \bar{K}^{0} & -\dfrac{\eta}{\sqrt{3}}+\sqrt{\dfrac{2}{3}} \eta^{\prime} \end{array} \right),\\ &V=\left(\begin{array}{ccc} \dfrac{\rho^{0}}{\sqrt{2}}+\dfrac{\omega}{\sqrt{2}} & \rho^+ & K^{*+} \\ \rho^- & -\dfrac{\rho^{0}}{\sqrt{2}}+\dfrac{\omega}{\sqrt{2}} & K^{* 0} \\ K^{*-} & \bar{K}^{* 0} & \phi \end{array}\right). \end{align} $
(16) Now, we can write the invariant mass distribution
$ M_{\rm inv}(K^0f_0(a_0)) $ in the decay$ B_0\to J/\psi K^0 f_0 $ as$ \begin{align} \frac{{\rm d}\Gamma_{B_0\to J/\psi K^0 f_0(a_0)}}{{\rm d}M_{\rm inv}(K^0 f_0(a_0))}=\frac{1}{(2\pi)^3}\frac{1}{4M_{B^0}^2}p^{\prime}_{J/\psi}\tilde{p}^{\prime}_{K^0}\overline{\sum}\sum|t|^2, \end{align} $
(17) We take the first diagram of Fig. 1 and the
$ B_0\to J/\psi K^0f_0 $ decay as an example and write down its amplitude as follows:$ \begin{aligned}[b] t=&\mathcal{C}\int \frac{{\rm d}^3 q}{(2\pi)^3}\frac{1}{8\omega_{K^0}\omega_\phi \omega_{K^0}^{\prime}}\frac{1}{k^0-\omega_{K^0}^{\prime}-\omega_{\phi}} \\ &\times \frac{1}{M_{\rm inv}(K^0f_0)+\omega_{K^0}+\omega_{K^0}^{\prime}-k^0} \\ &\times \frac{1}{M_{\rm inv}(K^0f_0)-\omega_{K^0}-\omega_{K^0}^{\prime}-k^0+i\dfrac{\Gamma_{K^0}}{2}} \\ &\times \left[ \frac{2M_{\rm inv}(K^0f_0)\omega_{K^0}+2k^0\omega_{K^0}^{\prime}}{M_{\rm inv}(K^0f_0)-\omega_{\phi}-\omega_{K^0}+i\dfrac{\Gamma_{\phi}}{2}+i\dfrac{\Gamma_{K^0}}{2}} \right. \\ &\left.-\frac{2(\omega_{K^0}+\omega_{K^0}^{\prime})(\omega_{K^0}+\omega_{K^0}^{\prime}+\omega_{\phi})}{M_{\rm inv}(K^0f_0)-\omega_{\phi}-\omega_{K^0}+i\dfrac{\Gamma_{\phi}}{2}+i\dfrac{\Gamma_{K^0}}{2}}\right]\\ &\times gg_{K^0\bar{K}^0f_0}p^{\prime}_{J/\psi}\tilde{p}^{\prime}_{K^0}\left( 2+\frac{\vec q \cdot \vec k}{{\vec k}^2} \right), \end{aligned} $
(18) where
$ \begin{aligned}[b] \omega_{K^0}^{\prime}=((\vec{P}-\vec{q}-\vec{k})^2+m_{K^0}^2)^{\frac{1}{2}},\end{aligned} $
$ \begin{aligned}[b] &\omega_{\phi}=((\vec{P}-\vec{q})^2+m_{\phi}^2)^{\frac{1}{2}},\\ &\omega_{K^0}=(\vec{q}^2+m_{K^0}^2)^{\frac{1}{2}},\\ &k^0=\frac{M_{\rm inv}^2(K^0f_0)+m_{K^0}^2-m_{f_0}^2}{2M_{\rm inv}(K^0f_0)},\\ &\left |\vec{k}\right |=\frac{\lambda^{\frac{1}{2}}(M^2_{\rm inv}(K^0f_0),m_{K^0}^2,m_{f_0}^2)}{2M_{\rm inv}(K^0f_0)}. \end{aligned} $
(19) The
$ f_0 \to K^0\bar{K}^0 $ and$ a_0\to K^0\bar{K}^0 $ vertices are obtained from the chiral unitary approach of Ref. [24] with$ g_{K^0\bar{K}^0f_0} $ =2567 MeV and$ g_{K^0\bar{K}^0a_0} $ =3875 MeV.$ p^{\prime}_{J/\psi} $ represents the momentum of the$ J/\psi $ in the$ B_0 $ rest frame, and$ \tilde{p}^{\prime}_{K^0}=|\vec{k}| $ represents the momentum of the$ K^0 $ in the$ K^0f_0(980) $ rest frame, where$ \begin{align} p^{\prime}_{J/\psi}&=\frac{\lambda^{\frac{1}{2}}(M_{B_0}^2,m_{J/\psi}^2,M_{\rm inv}^2(K^0f_0))}{2M_{B_0}}, \end{align} $
(20) $ \begin{align} \tilde{p}^{\prime}_{K^0}&=\frac{\lambda^{\frac{1}{2}}(M_{\rm inv}^2(K^0f_0),m_{K^0}^2,m_{f_0}^2)}{2M_{\rm inv}(K^0f_0)}, \end{align} $
(21) here,
$ t_T $ represents the loop amplitude:$ \begin{aligned}[b] t_T=&\int \frac{{\rm d}^3 q}{(2\pi)^3}\frac{1}{8\omega_{K^0}\omega_\phi \omega_{K^0}^{\prime}}\frac{1}{k^0-\omega_{K^0}^{\prime}-\omega_{\phi}} \\ &\times \frac{1}{M_{\rm inv}(K^0f_0)+\omega_{K^0}+\omega_{K^0}^{\prime}-k^0} \\ &\times \frac{1}{M_{\rm inv}(K^0f_0)-\omega_{K^0}-\omega_{K^0}^{\prime}-k^0+i\frac{\Gamma_{K^0}}{2}} \\ &\times \left[ \frac{2M_{\rm inv}(K^0f_0)\omega_{K^0}+2k^0\omega_{K^0}^{\prime}}{M_{\rm inv}(K^0f_0)-\omega_{\phi}-\omega_{K^0}+i\dfrac{\Gamma_{\phi}}{2}+i\dfrac{\Gamma_{K^0}}{2}} \right. \\ &\left.-\frac{2(\omega_{K^0}+\omega_{K^0}^{\prime})(\omega_{K^0}+\omega_{K^0}^{\prime}+\omega_{\phi})}{M_{\rm inv}(K^0f_0)-\omega_{\phi}-\omega_{K^0}+i\dfrac{\Gamma_{\phi}}{2}+i\dfrac{\Gamma_{K^0}}{2}}\right]\left( 2+\frac{\vec q \cdot \vec k}{{\vec k}^2} \right). \end{aligned} $
(22) We have
$ \begin{aligned}[b] &\frac{1}{\Gamma_{B_0}}\frac{{\rm d}\Gamma_{B_0\to J/\psi K^0 f_0}}{{\rm d}M_{\rm inv}(K^0f_0)}\\ =&\frac{1}{(2\pi)^3}\frac{1}{4M_{B^0}}p^{\prime}_{J/\psi}\tilde{p}^{\prime}_{K^0}\frac{\mathcal{C}^2}{\Gamma_{B_0}}\cdot g^2g^2_{K^0\bar{K}^0f_0}|\vec{k}|^2|t_T|^2\\ =&\frac{1}{(2\pi)^3}\frac{1}{4M_{B^0}}p^{\prime}_{J/\psi}\tilde{p}^{\prime 3}_{K^0}\frac{\mathcal{C}^2}{\Gamma_{B_0}}\cdot g^2g^2_{K^0\bar{K}^0f_0}|t_T|^2.\\ \end{aligned} $
(23) The case for
$ a_0 $ production is identical, except that$ g_{K^0\bar{K}^0f_0} $ is replaced with$ g_{K^0\bar{K}^0a_0} $ .The
$ K^0\bar{K}^0\to \pi^+\pi^- $ and$ K^0\bar{K}^0\to\pi^0\eta $ scattering was studied in detail in Refs. [43, 44] within the chiral unitary approach, where six channels were taken into account, including$ \pi^+\pi^- $ ,$ \pi^0\pi^0 $ ,$ K^+K^- $ ,$ K^0\bar{K}^0 $ ,$ \eta\eta $ , and$ \pi^0\eta $ . In the present study, we use this as input, and we shall see simultaneously both the$ f_0(980) $ (with I = 0) and$ a_0(980) $ (with I = 1) productions. Now, we can write the double differential mass distribution for the decay process$ B_0\to J/\psi K^0f_0(980)\to J/\psi K^0\pi^+\pi^- $ .For the case of
$ f_0(980) $ , we only have the decay$ f_0\to\pi^+\pi^- $ ; thus,$ \begin{aligned}[b] &\frac{{\rm d}^2\Gamma_{B_0\to J/\psi K^0f_0(980)\to J/\psi K^0\pi^+\pi^-}}{{\rm d}M_{\rm inv}(K^0f_0){\rm d}M_{\rm inv}(\pi^+\pi^-)}\\ = &\frac{1}{(2\pi)^5}\frac{1}{4M^2_{B_0}}p^{\prime\prime}_{J/\psi}\tilde p^{\prime\prime3}_{K_0}\tilde p^{\prime\prime}_{\pi^+}\sum\overline{\sum}|t^{\prime}|^2, \end{aligned} $
(24) where
$ p^{\prime\prime}_{J/\psi} $ represents the momentum of the$ J/\psi $ in the$ B_0 $ rest frame,$ \tilde p^{\prime\prime}_{K^0} $ represents the momentum of the$ K^0 $ in the$ K^0f_0(980) $ rest frame, and$ \tilde p^{\prime\prime}_{\pi} $ represents the momentum of the π in the$ \pi^+\pi^- $ rest frame:$ \begin{aligned}[b] &p^{\prime\prime}_{J/\psi}=\frac{\lambda^{\frac{1}{2}}(M_{B_0}^2,m_{J/\psi}^2,M_{\rm inv}^2(K^0f_0))}{2M_{B_0}},\\ &\tilde p^{\prime\prime}_{K_0}=\frac{\lambda^{\frac{1}{2}}(M_{\rm inv}^2(K^0f_0),m_{K^0}^2,M_{\rm inv}^2(\pi^+\pi^-))}{2M_{\rm inv}(K^0f_0)},\\ &\tilde p^{\prime\prime}_{\pi}=\frac{\lambda^{\frac{1}{2}}(M_{\rm inv}^2(\pi^+\pi^-),m_{\pi^+}^2,m_{\pi^-}^2)}{2M_{\rm inv}(\pi^+\pi^-)}. \end{aligned} $
(25) We can obtain the amplitude of
$ B_0\to J/\psi K^0f_0\to J/\psi K^0\pi^+\pi^- $ :$ \begin{align} \sum\overline{\sum}|t^{\prime}|^2=\mathcal{C}^2g^2|t_{K^0\bar{K}^0,\pi^+\pi^-}|^2|t_T|^2, \end{align} $
(26) where
$ \begin{align} t=[1-VG]^{-1}V, \end{align} $
(27) and the V matrix is taken from Ref. [24]. Then, we obtain the double differential branching ratio of the
$ B_0\to J/\psi K^0 f_0\to J/\psi K^0\pi^+\pi^- $ reaction:$ \begin{aligned}[b] \frac{1}{\Gamma}\frac{d^2\Gamma_{B_0\to J/\psi K^0f_0\to J/\psi K^0\pi^+\pi^-}}{dM(K^0f_0)dM_{\rm inv}(\pi^+\pi^-)}=&\frac{\mathcal{C}^2}{\Gamma_{B_0}}\frac{1}{(2\pi)^5}\frac{1}{4M^2_{B_0}}p^{\prime\prime}_{J/\psi}\tilde{p}^{\prime\prime3}_{K_0}\tilde{p}^{\prime\prime}_{\pi}\\ &\times g^2|t_{K^0\bar{K}^0,\pi^+\pi^-}|^2|t_T|^2. \end{aligned} $
(28) The case for
$ a_0 $ production is identical, except that$ t_{K^0\bar{K}^0,\pi^+\pi^-} $ is replaced with$ t_{K^0\bar{K}^0,\pi^0\eta} $ . -
Let us begin by showing in Fig. 3 the contribution of the triangle loop to the total amplitude and the triangle loop defined in Eq. (22). In order to satisfy the TS condition of Eq. (1), all intermediate particles must be on the shell; thus, the mass sum of
$ K^0\bar{K}^0 $ must be smaller than that of$ f_0(a_0) $ . If the mass of$ f_0(a_0) $ is too large, Eq. (1) will not be satisfied. There is hence a very narrow window of$ f_0(a_0) $ masses where the TS condition is exactly fulfilled, i.e., 995 to 999 MeV. The magnitude depends on the$ f_0(a_0) $ mass; it is independent of whether we have$ f_0 $ or$ a_0 $ , since the different couplings to$ K^0\bar{K}^0 $ have been factorized out of the integral of$ t_T $ . From the perspective of the above considerations, we plot the real and imaginary parts of$ t_T $ , as well as the absolute value with$ M_{\rm inv}(R) $ fixed at 986, 991, and 996 MeV. We can see that the bottom one has a clear peak, because 996 MeV is in the energy window. It can be observed that Re($ t_T $ ) has a peak around 1518 MeV, Im($ t_T $ ) has a peak around 1529 MeV, and there is a peak for$ |t_T| $ around 1520 MeV. As discussed in Refs. [45, 46], the peak of the imginary part is related to the TSs, while that of the real part is related to the$ K^0\phi $ threshold.$ |t_T|=\sqrt{{\rm Re}^2(t_T)+{\rm Im}^2(t_T)} $ is between$ {\rm Re}(t_T) $ and$ {\rm Im}(t_T) $ . Note that, at approximately 1520 MeV and above, the TS dominates the reaction.Figure 3. (color online) Triangle amplitude
$ t_T $ , as a function of$ M_{\rm inv}(K^0f_0/K^0a_0) $ for (a)$ m_{f_0(a_0)} $ = 986 MeV, (b)$ m_{f_0(a_0)} $ = 991 MeV, and (c)$ m_{f_0(a_0)} $ = 996 MeV.$ |t_T| $ , Re($ t_T $ ), and Im($ t_T $ ) are plotted using green, orange, and blue curves, respectively.Next, we show
$\dfrac{1}{\Gamma_{J/\psi}}\dfrac{{\rm d}\Gamma_{B^0\to J/\psi K^0f_0(a_0)}}{{\rm d}M_{\rm{inv}}(K^0f_0/K^0a_0)}$ . In Fig. 4, we plot Eq. (23) for the decay process$ B_0\to J/\psi K^0f_0(a_0) $ , and we see a peak around 1520 MeV. We can obtain the branching ratio of the 3-body decay process when we integrate over$ M_{\rm inv}(K^0R) $ ,Figure 4. (color online) Differential branching ratio
$\dfrac{1}{\Gamma_{B_0}}\dfrac{{\rm d}\Gamma_{B_0\to J/\psi K^0 f_0}}{{\rm d}M_{\rm inv}(K^0f_0)}$ described in Eq. (23) as a function of$ M_{\rm inv}(K^0f_0/K^0a_0) $ $ \begin{align} {\rm Br}(B_0\to J/\psi K^0f_0(a_0))=2.45\times10^{-6}. \end{align} $
(29) In the upper panel of Fig. 5, we plot Eq. (24) for the
$ B_0\to J/\psi K^0\pi^+\pi^- $ decay, and similarly, in the lower panel of Fig. 5, we plot it for the$ B_0\to J/\psi K^0\pi^0\eta $ decay as a function of$ M_{\rm inv}(R) $ . In both figures, we fix$ M_{\rm inv}(K^0R) $ = 1500, 1520, and 1540 MeV and vary$ M_{\rm inv}(R) $ . We can see that the distribution with the highest strength is near$ M_{\rm inv}(K^0R) $ = 1520 MeV. We also observe a strong peak when$ M_{\rm inv}(\pi^+\pi^-) $ is around 980 MeV in the upper panel of Fig. 5. Similar results are shown in the lower panel of Fig. 5. We see that most of the contribution to the width Γ comes from$ M_{\rm inv}(K^0R) $ =$ M_R $ , and we have strong contributions for$ M_{\rm inv}(\pi^+\pi^-)\in $ [500 MeV, 990 MeV] and$ M_{\rm inv}(\pi^0\eta)\in $ [800 MeV, 990 MeV]. Therefore, when we calculate the mass distribution$\dfrac{{\rm d}\Gamma}{{\rm d}M_{\rm inv}(K^0R)}$ , we restrict the integral to the limits already mentioned and perform the integration.Figure 5. (color online) (a)
$\dfrac{{\rm d}^2\Gamma_{B_0\to J/\psi K^0f_0(980)\to\pi^+\pi^-}}{{\rm d}M_{\rm inv}(K^0f_0){\rm d}M_{\rm inv}(\pi^+\pi^-)}$ as a function of$ M_{\rm inv}(\pi^+\pi^-) $ . (b)$\dfrac{{\rm d}^2\Gamma_{B_0\to J/\psi K^0a_0(980)\to J/\psi K^0\pi^0\eta}}{{\rm d}M_{\rm inv}(K^0a_0){\rm d}M_{\rm inv}(\pi^0\eta)}$ as a function of$ M_{\rm inv}(\pi^0\eta) $ . The blue, orange, and green curves are obtained by setting$ M_{\rm inv}(K^0f_0)= $ 1500, 1520, and 1540 MeV, respectively.$ \begin{aligned}[b] &\frac{1}{\Gamma_{B_0}} \frac{{\rm d} \Gamma_{B_0\to J/\psi K^0 f_0(980) \to J/\psi K^0\pi^+\pi^-}}{{\rm d} M_{\mathrm{inv}}(K^0f_0)} \\ =&\frac{1}{\Gamma_{J/\psi}} \int_{500 \mathrm{MeV}}^{990 \mathrm{MeV}} {\rm d} M_{\mathrm{inv}}(\pi^+\pi^-) \\ & \times \frac{{\rm d}^{2} \Gamma_{B_0\to J/\psi K^0 f_0(980) \to J/\psi K^0\pi^+\pi^-}}{{\rm d} M_{\mathrm{inv}}(K^0f_0) {\rm d} M_{\mathrm{inv}}(\pi^+\pi^-)}, \end{aligned} $
(30) $ \begin{aligned}[b] &\frac{1}{\Gamma_{B_0}} \frac{{\rm d} \Gamma_{B_0\to J/\psi K^0 a_0(980) \to J/\psi K^0\pi^0\eta}}{{\rm d} M_{\mathrm{inv}}(K^0a_0)} \\ =&\frac{1}{\Gamma_{J/\psi}} \int_{800 \mathrm{MeV}}^{990 \mathrm{MeV}} {\rm d} M_{\mathrm{inv}}(\pi^0\eta) \\ &\times \frac{{\rm d}^{2} \Gamma_{B_0\to J/\psi K^0 a_0(980) \to J/\psi K^0\pi^0\eta}}{{\rm d} M_{\mathrm{inv}}(K^0a_0) {\rm d} M_{\mathrm{inv}}(\pi^0\eta)}. \end{aligned} $
(31) We show Eq. (24) for both
$ B_0\to J/\psi K^0\pi^+\pi^- $ and$ B_0\to J/\psi K^0\pi^0\eta $ . When we integrate over$ M_{\rm inv}(R) $ , we obtain$\dfrac{{\rm d}\Gamma}{{\rm d}M_{\rm inv}(K^0R)}$ , which we show in Fig. 6. We see a clear peak of the distribution around 1520 MeV, for$ f_0 $ and$ a_0 $ production. At the same time, we observe that the peak of$ M_{\rm inv}(K^0a_0) $ is significantly lower than the peak of$ M_{\rm inv}(K^0f_0) $ .Figure 6. (color online) (a) Differential branching ratio
$\dfrac{{\rm d}\Gamma_{B_0\to J/\psi K^0f_0(980)\to J/\psi K^0\pi^+\pi^-}}{{\rm d}M_{\rm inv}(K^0f_0)}$ . (b) Branching ratio$\dfrac{{\rm d}\Gamma_{B_0\to J/\psi K^0f_0(980)\to J/\psi K^0\pi^+\pi^-}}{{\rm d}M_{\rm inv}(K^0a_0)}$ described in Eq. (24) as a function of$ M_{\rm inv}(K^0f_0/K^0a_0) $ By integrating
$\dfrac{{\rm d}\Gamma}{{\rm d}M_{\rm inv}(K^0f_0)}$ and$\dfrac{{\rm d}\Gamma}{{\rm d}M_{\rm inv}(K^0a_0)}$ over the$ M_{\rm inv}(K^0f_0) $ ($ M_{\rm inv}(K^0a_0) $ ) masses in Fig. 6, we obtain the branching fractions:$ \begin{align} &{\rm Br}(B_0\to J/\psi K^0f_0(980)\to J/\psi K^0\pi^+\pi^-)=7.67\times10^{-7},\\ &{\rm Br}(B_0\to J/\psi K^0a_0(980)\to J/\psi K^0\pi^0\eta)=1.42\times10^{-7}. \end{align} $
(32)
Triangle mechanism in the decay process $ {\boldsymbol B_{\bf 0}{\bf \to}{\boldsymbol J}{\bf /}\boldsymbol\psi {\boldsymbol K}^{\bf 0} {\boldsymbol f}_{\bf 0}{\bf (980)}({\boldsymbol a}_{\bf 0}{\bf(980))} }$
- Received Date: 2023-03-16
- Available Online: 2023-07-15
Abstract: The role of the triangle mechanism in the decay processes